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ikadub [295]
3 years ago
15

Comment Both propane and benzene are hydrocarbons. As a rule,

Chemistry
1 answer:
kozerog [31]3 years ago
5 0

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

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Write the balanced chemical equation for the dissociation reaction of thiosulfuric acid.
mafiozo [28]

Answer:

f(x)=3x^2

Explanation:

5 0
3 years ago
24 g of magnesium were burned in oxygen. The compound formed had a mass of 40 g. Explain why the mass had gone up.
DIA [1.3K]

Answer :

According to the law of conservation of mass, the mass of reactants must be equal to the mass of products.

The balanced chemical reaction is,

Mg+\frac{1}{2}O_2\rightarrow MgO

As we know that the molar mass of magnesium is 24 g/mole, the molar mass of O_2 is 32 g/mole and the molar mass of magnesium oxide is 40 g/mole.

From the given balanced reaction, we conclude that

As, 1 mole of magnesium react \frac{1}{2} mole of oxygen to give 1 mole of magnesium oxide.

So, the mass of Mg is 24 g, the mass of O_2=\frac{1}{2}\times 32=16g and the mass of MgO is 40 g.

That means 24 g of Mg react with 16 g O_2 to give 40 g of MgO.

8 0
3 years ago
How many significant figures are in 0.000128?
san4es73 [151]
Three sig figs, the leading zeros are not sig figs
6 0
3 years ago
How many moles of carbon dioxide (CO2) are in a 22 gram sample?
Anvisha [2.4K]

Answer:

0.5 mole

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C=12u

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6 0
3 years ago
g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint
diamong [38]

Answer:

The value  is   L  =  0.985 \pm 0.00801 \  g

Explanation:

From the question we are told that

  The  molar mass of CaCO_3 is  MW  =  100.085 \  g/mol

   The  total mass is  m_g  = 12.3 \ g

   The uncertainty of the total mass is \Delta g  = 0.1

Generally the molar weight of calcium is M_c  =  40 g/mol

 The percentage of calcium in calcite is mathematically represented as

          C =  \frac{40.07}{100.085} * 100

          C =  40.03 \%

Generally the mass of each sample is mathematically represented as

     m=  \frac{m_g}{5}

     m=  \frac{12.3}{5}

     m= 2.46 \  g

Generally mass of calcium present in a single sample is mathematically represented as

        m_c = 2.46 *  \frac{40.04}{100}

       m_c = 0.985 \  g

The  uncertainty of  mass of a single sample is mathematically represented as

      k  =  \frac{\Delta g }{5}

        k  =  \frac{0.1 }{5}

       k  =  0.02\  g

The  uncertainty of  mass of calcium in a single sample is mathematically represent

         G  =  \frac{0.02 *  40.04}{ 100}

          G  =  0.00801 \  g

Generally the average mass of calcium in each sample is  

          L  =  0.985 \pm 0.00801

6 0
3 years ago
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